Problem: A regular polygon has interior angles of 120 degrees. How many sides does the polygon have?
Solution: Let $n$ be the number of sides in the polygon.  The sum of the interior angles in any $n$-sided polygon is $180(n-2)$ degrees.  Since each angle in the given polygon measures $120^\circ$, the sum of the interior angles of this polygon is also $120n$.  Therefore, we must have  \[180(n-2) = 120n.\] Expanding the left side gives $180n - 360 = 120n$, so $60n = 360$ and $n = \boxed{6}$.

We might also have noted that each exterior angle of the given polygon measures $180^\circ - 120^\circ = 60^\circ$.  The exterior angles of a polygon sum to $360^\circ$, so there must be $\frac{360^\circ}{60^\circ} = 6$ of them in the polygon.